Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - Review - Exercises - Page 1189: 3

Answer

$\int_Cyzcosxds=6\sqrt {10}$

Work Step by Step

$ds=\sqrt {(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}}dt$ $ds=\sqrt {{1}^{2}+({-3sint})^{2}+({3cost})^{2}}dt$ $ds=\sqrt {1+9}dt$ $ds=\sqrt {10}dt$ Now, $\int_Cyzcosxds=\int _{0}^{\pi}(3cost)(3sint)cos(t)\sqrt {10}dt$ $=9\sqrt {10}\int _{0}^{\pi}cos^{2}tsintdt$ Substitute $cost=u$ and $-sintdt=du$ Limits of integration will change from $\int _{0}^{\pi}$ to$\int _{1}^{-1}$. $=-9\sqrt {10}\int _{1}^{-1}u^{2}du$ $=-9\sqrt {10}[\frac{u^{3}}{3}] _{1}^{-1}$ $=-9\sqrt {10}[\frac{-1}{3}-\frac{1}{3}]$ Hence, $\int_Cyzcosxds=6\sqrt {10}$
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